growth and decay

Question

1. A biologist grows a colony of a colony of a certain kind of bacteria.  It is found experimentally that N= N(0)3^t , where N represents the number of bacteria present at the end of t days.  N(0) is the number of bacteria present at the start of the experiment.  Suppose that there are 153,000 bacteria present at the end of 2 days.

 

1. How many bacteria were present at the start of the experiment?

 

1. How many bacteria are present at the end of 4 days?

 

1. At the end of how many days are 459,000 bacteria present?

 

1. Justify your solution.

Answer 1

N=153,000=N(0)*3^2=9N(0), so N(0)=153000/9=17,000.

After 4 days there are 9*153000=1,377,000.

The growth to 459,000 is 459000/17000=27=3^3, corresponding to t=3 days.